9
$\begingroup$

Given a small category $\mathbb C$, we can form the free cocompletion $\mathbf y : \mathbb C \to \mathcal P(\mathbb C)$ and the free completion $\mathbf y^\circ : \mathbb C \to \mathcal P^\circ(\mathbb C)$. These constructions factor through two others:

  1. The free bicompletion $\mathbf b : \mathbb C \to \mathcal B(\mathbb C)$.
  2. The Isbell envelope $\mathbf i :\mathbb C \to \mathcal I(\mathbb C)$.

free bicompletion and Isbell envelope

Both constructions have universal properties: the former freely adds limits and colimits; whilst the latter freely constructs a cylinder factorisation system.

However, it is not clear to me exactly how the two relate. May one construction be expressed in terms of the other (e.g. is there a factorisation of free bicompletion through the Isbell envelope)? If not, is there some other relationship?

Improve this question
$\endgroup$
2
  • 1
    $\begingroup$ I'd like to know ! When $\mathbb{C}$ is small the Isbell completion is complete and cocomplete correct ? So we always have a limite & colimit preserving functor $\mathcal{B}(\mathbb{C}) \to \mathcal{I}(\mathbb{C})$. Do you know if this functor has right and left adjoints ? I suspect not, but that's the only way I can think to send the Isbell completion to the Bi-completion. $\endgroup$
    – Simon Henry
    Commented Oct 27, 2020 at 11:14
  • $\begingroup$ My understanding is that it is bicomplete (which follows from the definition of CFS). However, I should note that there is an issue with terminology, in that you'll find it stated (e.g. in this answer) that the Isbell envelope may not have small limits or colimits: but this instead refers to the construction arising from the Isbell adjunction (the distinction is discussed here). I don't know whether the functor you mention has adjoints, either. $\endgroup$
    – varkor
    Commented Oct 27, 2020 at 11:57

0

Reset to default

You must log in to answer this question.